Determination of the number of $R$ submodules where $R= \mathbb{Z}/4\mathbb{Z}$ Suppose $R=\mathbb{Z}/4\mathbb{Z}$.  
i) How many R-submodules $M= Rx \subset R^{2} \ (x\in R^{2}) $ are there?
ii) How many equivalence classes are isomorphic to M? 
i) definition of a submodule: Let M be a R-module an $L\subset M$, then L is a R-submodule of M if L itself is a R-module respectively to the  operation on M. 
every ideal I of R is a R-submodule of R. The ideals of R are the equivalence classes $\overline{0,1,2,3}$, so the submodules of M in this case are : 0x,1x,2x,3x. From this it would also follow that the number of equivalence classes is equal to the number of the submodules (ii).
Is this reasoning alright?
 A: $\newcommand{\Z}{\mathbf{Z}}$Let's make sure we know how to deal with $R$, first! Maybe it's helpful to note that we're really asking questions about abelian groups here: an $R$-submodule of $R$ is the same as a $\Z$-submodule of $R$. In any event, the submodules of $R$ correspond to subgroups of $\Z$ which contain $4\Z$, and there are three of these.
Now you want to find the cyclic $R$-submodules of $R \times R$. This module only has $16$ elements, so it isn't that hard to just check each one [Say I take $(1, 2) \in R \times R$. Then this generates $\{(1, 2), (2, 0), (3, 2), (0, 0)\}$, which is isomorphic to $R$]. But some observations can make the job easier.
Note that the order of $(a, b)$ is the l.c.m. of the orders of $a$ and $b$. If I want to find the cyclic submodules of order $4$, i.e. the submodules isomorphic to $R$, then I want $a$ or $b$ of order $4$. So I want pairs like $(1, b), (3, b)$, and the mirror versions of these. By my count, there are $12$ such pairs. Here we have to be careful: each of these elements will generate a submodule that contains two elements of order $4$, so in the end there are only $6$ submodules of this sort.
